I have a 3D cube, 8 3D point cordinates, whose center is $(0,0,0)$, and is rotationally distorted. I want to calculate the angles, $R_x, R_y, R_z$ for rotation to rectificate the cube.

3D coordinates, $(x,y,z)$'s, of a sample cube:

 p0 = (0,4482,   0,6137,   0,4153)
 p1 = (0,1041,   0,6839,  -0,5210)
 p2 = (0,7758,  -0,3119,   0,2255)
 p3 = (0,4317,  -0,2417,  -0,7108)
 p4 = (-0,4317,  0,2417,   0,7108)
 p5 = (-0,7758,  0,3119,  -0,2255)
 p6 = (-0,1041,  -0,6839,  0,5210)
 p7 = (-0,4482,  -0,6137, -0,4153)

which looks something like:

enter image description here

(the camera is at the point $(0,0,2)$ and looks to the point $(0,0,0)$)

I created 3 vectors using coordinates of the Cube, $V_x', V_y', V_z'$, $p_0 - p_1, p_0 - p_2, p_0-p_4$ these 3 vectors have some angles with the real world axis', $V_x, V_y, V_z$.

I know how to use dot product, however, using classical dot product does not give me the true rotation angles, $R_x, R_y, R_z$, and indeed I know the ordering of the rotation is important, (i.e., rotate X axis 30; rotate Y axis 40; is not the same as rotate Y axis 40; rotate X axis 30;).

My question is using the coordinate system of the cube, $V_x', V_y', V_z'$ and using the real coordinate system, $V_x, V_y, V_z$, $[1,0,0], [0,1,0],[0,0,1]$, how can I come up with the $R_x, R_y, R_z$ angles such that when I rotate all the points of the cube using $R_x, R_y, R_z$, respectively, the cube will be rectified (i.e., $V_x$ and $V_x'$, $V_y$ and $V_y'$, $V_z$ and $V_z'$ will be one after the other) ?

Or if my current approach is wrong, how can I come up with the angles, $R_x, R_y, R_z$, using $p_0, p_1, \ldots, p_7$ ?

  • 1
    $\begingroup$ Can you get away with just forming a rotation matrix instead of having to recover the Euler angles? If so, the $3\times3$ matrix with $V_x'$, $V_y'$, and $V_z'$ as rows will transform your cube into an axis-aligned one. $\endgroup$ – user856 Aug 20 '12 at 18:54
  • $\begingroup$ @RahulNarain I need to get the angles as well. I thought it should be easy, isn't it? $\endgroup$ – Sait Aug 20 '12 at 19:45
  • 1
    $\begingroup$ The thing about Euler angles is that they are order-dependent, so if I give you a formula that assumes one convention and you're using a different one, it doesn't help you... You can look at Wikipedia's formula for converting a rotation matrix to angles in the $ZXZ$ convention, compare it with the list of rotation matrices for different conventions, and try to adapt the formula for the convention you're using. $\endgroup$ – user856 Aug 20 '12 at 20:08
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    $\begingroup$ Since you haven't told us which ordering of rotations you're using, that's about all I can say at this point. $\endgroup$ – user856 Aug 20 '12 at 20:10
  • 1
    $\begingroup$ It's just the matrix in my first comment. $\endgroup$ – user856 Aug 20 '12 at 22:33

Since I'm typing in a full answer, here's a little linear algebra for free. :)

If your original coordinate axes have been linearly transformed to the vectors $V_x'$, $V_y'$, and $V_z'$, the matrix which performed this transformation must have been $$R = \begin{bmatrix}V_x' & V_y' & V_z'\end{bmatrix},$$ i.e. the $3\times3$ matrix whose columns are $V_x'$, $V_y'$, and $V_z'$. (Try multiplying this matrix with the three axis-aligned unit vectors to see why.) The matrix you need to undo that transformation is its inverse, $R^{-1}$. Since in your case $R$ is a rotation matrix, $V_x'$, $V_y'$ and $V_z'$ must be of unit length and mutually orthogonal. This also means that $R^{-1}$ is the same as the transpose, $R^T$, which is the matrix whose rows are $V_x'$, $V_y'$, and $V_z'$: $$R^{-1} = R^T = \begin{bmatrix}V_x'^T \\ V_y'^T \\ V_z'^T\end{bmatrix}.$$

So now you have a rotation matrix $R^{T}$ which transforms your cube back to being axis-aligned. But you want the corresponding Euler angles, say in the order $ZXZ$. Wikipedia gives the matrix corresponding to rotation angles $\theta_1$, $\theta_2$, $\theta_3$ as $$A = R_Z(\theta_1) R_X(\theta_2) R_Z(\theta_3) = \begin{bmatrix} c_1 c_3 - c_2 s_1 s_3 & - c_1 s_3 - c_2 c_3 s_1 & s_1 s_2 \\ c_3 s_1 + c_1 c_2 s_3 & c_1 c_2 c_3 - s_1 s_3 & - c_1 s_2 \\ s_2 s_3 & c_3 s_2 & c_2 \end{bmatrix},$$ where $c_i$ and $s_i$ mean $\cos\theta_i$ and $\sin\theta_i$ respectively. So we can see that $$\begin{align} a_{13}/a_{23} &= -\tan\theta_1, \\ a_{33} &= \cos\theta_2, \\ a_{31}/a_{32} &= \tan\theta_3. \\ \end{align}$$ Of course, $A$ is nothing but $R^T$, so we already know the values of its entries, and we can recover $\theta_1$, $\theta_2$, and $\theta_3$.

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